$\int (-4 x^3 -8 x^2 -4)\,dx=$ $+C$
Answer: We can use the sum rule and the constant multiple rule for indefinite integrals: $\begin{aligned} &\int [f(x)+g(x)]dx=\int f(x)\,dx+\int g(x)\,dx \\\\\\ &\int k\cdot f(x)= k\cdot\int f(x)\,dx \end{aligned}$ Using the sum and the constant multiple rules, we can rewrite our integral as follows: $\int (-4 x^3 -8 x^2 -4)\,dx= -4\int x^3\,dx -8\int x^2\,dx -4\int 1\,dx$ Now we can find each indefinite integral using the reverse power rule: $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ Note: we can only use the reverse power rule because $n \neq -1$. $\begin{aligned} &\phantom{=}\int (-4 x^3 -8 x^2 -4)\,dx \\\\ &= -4\int x^3\,dx -8\int x^2\,dx -4\int 1\,dx \\\\ &=-4 \dfrac{x^4}{4} -8\dfrac{x^3}{3} -4\dfrac{x^1}{1}+C \\\\ &=- x^4 -\dfrac{8}{3} x^3 -4 x+C \end{aligned}$ In conclusion, $\int (-4 x^3 -8 x^2 -4)\,dx=- x^4 -\dfrac{8}{3} x^3 -4 x+C$